In this video I go over how to decompose rational functions that have repeating factors in the denominator. The method is similar to that shown in my earlier video on general techniques of partial fraction decomposition but we need to account for all the different possible partial fraction cases for when the factors repeat. Solving for the coefficients of partial fractions is more complicated for these cases so it is usually easier to select values of x before solving for these coefficients.
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Partial Fraction Decomposition: Repeating Factors
In my earlier video I went over general techniques for partial fraction decomposition and showed how to decompose functions with unique and linear factors in the denominator:
But what if the factors repeat and/or are non-linear? Consider the rational function with repeating factor (x - 1):
Now in this case we can't just assume that the partial fractions contain only x or (x-1) in the denominator. This is because we can possibly have a (x-1)2 or (x-1)3 in the denominator of the partial fractions and still add up to make our function.
Thus we can still use the same method we went over in the previous video but this time we need to account for all possibilities of partial fractions:
In this example, one of the coefficients turned out to be zero but this is not always the case and sometimes the coefficients are fractions too (i.e. 1/3, 2/5, etc).